A binomial distribution can be considered as the probability of a success or failure outcome in a repeated trial or experiment. The binomial distribution is a sort of probability distribution with two possible outcomes (the prefix “bi” signifies “two”). Binomial distributions can be encountered in a wide variety of situations in everyday life.

Tossing a coin, for example, has only two outcomes: heads or tails, and passing a test has only two outcomes: pass or fail. When a new drug is released to treat a condition, it either cures the ailment (success) or does not (failure). You either win money, or you do not if you buy a lottery ticket. In this article, we will provide detailed information on the binomial distribution. Continue reading to learn more about binomial distribution!

Binomial Distribution: Overview

The binomial distribution is a discrete probability distribution in probability theory and statistics with only two possible outcomes in an experiment: success or failure, for a finite number of trials. If we toss a coin, for example, there are only two conceivable outcomes: heads or tails, and if we take a test, there are only two possible outcomes: pass or fail.

In a binomial distribution, the parameters \(n\) and \(p\) are considered. The variable \(n\) indicates the number of times the experiment is repeated, whereas the variable \(p\) indicates the probability of each event.

Consider the case if you wanted to determine the probability of getting a three on a die. The chance of getting a three on any throw is \(\frac{1}{6}\) if you throw a die \(20\) times. If you throw \(20\) times, you will get a binomial distribution of \(\left( {n = 20,p = \frac{1}{6}} \right).\) FAILURE would be “roll anything else,” while SUCCESS would be “roll a three.” The binomial distribution would become \(\left( {n = 20,p = \frac{1}{2}} \right)\) if the outcome was the probability of the dice landing on an even number. It is because tossing an even number has a half probability.

A set of Bernoulli trials is a Bernoulli distribution. Each Bernoulli trial has only one possible outcome: success or failure (\(S\) or \(F\)). The chance of success in each trial is the same, \(P\left( S \right) = p.P\left( F \right) = 1 – p.\) Finally, even if you know the results of past Bernoulli trials, the probability of success does not alter from trial to trial.

Tossing a coin, for example, is a Bernoulli trial because each trial can only have one of two outcomes (heads or tails), each success has the same probability (flipping a head has a probability of \(0.5\)), and the results of one trial have no bearing on the results of another. The Bernoulli distribution is a variant of the binomial distribution, with \(n=1\) as the number of trials.

Binomial Distribution Formula

A random variable \(X\) which takes values \(0,1,2,…, n\) is said to follow binomial distribution if its probability distribution function is given by
\(P\left( {X = r} \right)\, = \,{}^n{C_r}{p^r}{q^{n – r}},\,r = 0,\,1,\,2,\,…..,\,n\) where \(p,q > 0\) such that \(p + q = 1.\)

Here, 
\(n=\) No of trials
\(p=\) probability of success of each trial
\(q=\) probability of failure of each trial

The notation \(X \sim B(n,p)\) is generally used to denote that the random variable \(X\) follows a binomial distribution with parameters \(n\) and \(p.\)
We have
\(P(X = 0) + P(X = 1) + \ldots \ldots + P(X = n)\)
\({ = ^n}{C_0}{p^0}{q^{n – 0}}{ + ^n}{C_1}{p^1}{q^{n – 1}} + \ldots \ldots .{ + ^n}{C_n}{p^n}{q^{n – n}} = {(q + p)^n} = {1^n} = 1\)

Now probability of:

(a) Occurrence of the event exactly \(r\) times \(P\left( {X = r} \right)\, = \,{}^n{C_r}{q^{n – r}}{p^r}.\)
(b) Occurrence of the event at least \(r\) times \(P\left( {X \ge r} \right)\, = \,{}^n{C_r}{q^{n – r}}{p^r} + ….. + {p^n}\, = \,\sum\limits_{x = r}^n {{}^n{C_x}{p^x}{q^{n – x}}} \)
(c) Occurrence of the event at the most \(r\) times
\(P\left( {0 \le X \le r} \right)\, = \,{q^n} + {}^n{C_1}{q^{n – 1}}p + ….. + {}^n{C_r}{q^{n – 1}}{p^r}\, = \,\sum\limits_{x = 0}^r {{}^n{C_x}{p^x}{q^{n – x}}} \)

Properties of Binomial Distribution

The various properties of the binomial distribution are as follows:

1. The two possible outcomes of each independent trial are success or failure.
2. There are a fixed number of \(n\) times repeated trials or a fixed number of independent trials.
3. Each trial’s probability of success or failure is different.
4. Out of \(n\) independent trials, only the number of successes is calculated.
5. Every trial is independent of the others, which implies that the result of one trial does not affect the outcome of another.

Mean and Variance of Binomial Distribution

Mean and Variance of Binomial distribution are calculated from the following formula:

Mean \( = \mu  = np\)
Variance \( = {\sigma ^2} = npq\)
Where,
\(n = \) No of trials
\(p = \) probability of success of each trial
\(q = \) probability of failure of each trial

Solved Examples on Binomial Distribution

Q.1. The probability of a shooter hitting a target is \(\frac{3}{4}.\) How many minimum numbers of times must he/she fire so that the probability of hitting the target at least once is more than \(0.99\)?
Ans: Given that probability of the shooter hitting a target is \(\frac{3}{4}\)
Let the shooter fire \(n\) times and let \(X\) denote the number of times the shooter hits the target.
Then, \(X\) follows a binomial distribution with \(p = \frac{3}{4}\) and \(q = \frac{1}{4}\) such that
\(P\left( {X = r} \right)\, = \,{}^n{C_r}{\left( {\frac{3}{4}} \right)^r}{\left( {\frac{1}{4}} \right)^{n – r}}\)
\( \Rightarrow P\left( {X = r} \right)\, = \,{}^n{C_r}\frac{{{3^r}}}{{{4^n}}}\)
It is given that
\(P(X \ge 1) > 0.99\)
\( \Rightarrow 1 – P(X = 0) > 0.99\)
\( \Rightarrow 1 – \frac{1}{{{4^n}}} > 0.99\)
\( \Rightarrow \frac{1}{{{4^n}}} < 0.01\)
\( \Rightarrow {4^n} > \frac{1}{{0.01}}\)
\( \Rightarrow {4^n} > 100\)
The least value of \(n\) satisfying this inequality is \(4.\) Hence, the shooter must fire at least \(4\) times.

Q.2. A die is thrown three times. Let \(X\) be ‘the number of two’s seen’. Find the expectation (Mean) of \(X.\)
Ans: Let \(p\) be the probability of getting \(2\) when a dice is thrown. \( \Rightarrow p = \frac{1}{6}\)
Clearly, \(X\) follows a binomial distribution with \(n = 3,p = \frac{1}{6}\)
Expectation \( = E(X) = np = 3 \times \frac{1}{6} = \frac{1}{2}\)

Q.3. A die is tossed twice. If getting an even number on a toss is a success. Find the variance of the number of successes.
Ans: Let \(p\) be the probability of getting an even number on the toss when a dice is thrown.
Let \(q\) be the probability of not getting an even number on the toss when a dice is thrown.
Then \(p = \frac{3}{6} = \frac{1}{2}\) and \(q = 1 – p = \frac{1}{2}\)
Clearly, \(X\) follows a binomial distribution with \(n = 2,p = \frac{1}{2}\)
variance \( = npq = 2 \times \frac{1}{2} \times \frac{1}{2} = \frac{1}{2}\)

Q.4. A fair coin is tossed a fixed number of times. If the probability of getting \(7\) heads is equal to that of getting \(9\) heads, find the probability of getting two heads
Ans: Let \(X\) denote the number of heads in a fixed number of tosses of a coin.
Then, \(X\) is a binomial variate with parameters \(n\) and \(p = \frac{1}{2}\)
We know that \(P\left( {x = r} \right)\, = \,{}^n{C_r}{p^r}{q^{n – r}}\)
Given that \(P\left( {X = 7} \right) = P\left( {X = 9} \right)\)
Also \(p = q = 0.5\)
\(P\left( {X = r} \right)\, = \,{}^n{C_r}{\left( {0.5} \right)^r}{\left( {0.5} \right)^{n – r}} = {}^n{C_r}{\left( {0.5} \right)^n}\)
\(\therefore P\left( {X = 7} \right)\, = \,{}^n{C_7}{\left( {0.5} \right)^n}\) and \(P\left( {X = 9} \right)\, = \,{}^n{C_9}{\left( {0.5} \right)^n}\)
It is given that \(P(X = 7) = P(X = 9)\)
\(\therefore {}^n{C_7}{\left( {0.5} \right)^n} = {}^n{C_9}{\left( {0.5} \right)^n} \Rightarrow \frac{{n!}}{{7!\left( {n – 7} \right)!}} = \frac{{n!}}{{9!\left( {n – 9} \right)!}} \Rightarrow 9 \times 8 = \left( {n – 7} \right)\left( {n – 8} \right)\)
\( \Rightarrow {n^2} – 8n – 7n + 56 = 72 \Rightarrow {n^2} – 15n – 16 = 0\)
\( \Rightarrow (n + 1)(n – 16) = 0 \Rightarrow n =\, – 1\)or \(n = 16\)
\( \Rightarrow n = \, – 1\) (Not possible as \(n\) denotes the number of tosses of a coin), therefore \(n=16\)
Hence \(P\left( {X = 2} \right)\, = \,{}^{16}{C_2}{\left( {0.5} \right)^{16}}\)
\( = \frac{{16 \times 15}}{2} \times \frac{1}{{{2^{16}}}} = \frac{{15}}{{{2^{13}}}}\)

Q.5. A fair die is tossed eight times. Find the probability that a third six is observed in the eighth throw.
Ans:
Let \(p\) be the probability of obtaining a six in a single throw of the die.
Then, \(p = \frac{1}{6}\) and \(q = 1 – \frac{1}{6} = \frac{5}{6}\)
Obtaining a third six in the eighth throw of the die means that in the first seven throws, there are \(2\) sixes, and the third six is obtained in the eighth throw.
Therefore, required probability \(=P\)( Getting \(2\) sixes in the first seven throws) \(P\) (Getting six in the eighth throw)
\( = \left( {{}^7{C_2}{p^2}{q^{7 – 2}}} \right)p\, = \,{}^7{C_2}{\left( {\frac{1}{6}} \right)^2}{\left( {\frac{5}{6}} \right)^5} \times \frac{1}{6} = \frac{{{}^7{C_2} \times {5^5}}}{{{6^8}}}\)

Summary

In this article, we defined binomial distribution, its formula and solved examples. We understood that the binomial distribution is a discrete probability distribution employed when a random variable has just two possible outcomes: success or failure. Both success and failure are mutually exclusive, i.e., they cannot coexist. A finite number of trials, \(n,\) is assumed in the binomial distribution.

Each trial is distinct from the previous one. This means that the chance of success \(p\) remains constant from trial to trial. The probability of failure \(q\) is equal to \(1–p.\) Thus, success and failure probabilities are complementary.

Learn About Binomial Theorem Here

FAQs on Binomial Distribution

Q.1. Define binomial distribution and why is it called binomial?
Ans: The binomial distribution is a discrete probability distribution in probability theory and statistics that has only two possible outcomes in an experiment: success or failure for a finite number of trials. It is called the binomial because each trial outcome is two: success or failure.

Q.2. How do you find the \(n\) and \(p\) of a binomial distribution?
Ans: If the mean and variance of the binomial distribution are known to us, then \(q=1-p\) is found out by just dividing variance by the mean. Hence from \(q\) we can find out \(p\) easily.

Q.3. What are the binomial distribution parameters?
Ans: In a binomial distribution, the parameters \(n\) and \(p\) are considered. The variable \(n\) indicates the number of times the experiment is repeated, whereas the variable \(p\) indicates the probability of each event.

Q.4. What is binomial distribution?
Ans: The probability of observing \(X\) successes in \(n\) trials is calculated using the binomial distribution, with the probability of success on a single trial given by \(p.\) For all trials, the binomial distribution assumes that \(p\) is constant.
The formula for the probability distribution function is given by
\(P\left( {X = r} \right)\, = \,{}^n{C_r}{p^r}{q^{n – r}},\,r = 0,\,1,\,2,\,…..,\,n\) where \(p,q > 0\) such that \(p + q = 1\)
Here,
\(n=\) No of trials
\(p=\) probability of success of each trial
\(q=\) probability of failure of each trial

Q.5. What is a binomial in math?
Ans: In math, a binomial is a two-term polynomial with variables, coefficients, exponents, and constants.